Matrices fixing a subspace and the quotient by it Let $V$ be a finite-dimensional $\mathbb{C}$-vector space, $W$ be a $\mathbb{C}$-subspace. Is it true an invertible matrix that fixes $W$ and $V/W$ but not $V$ generates an infinite cyclic group? I can verify this if the dimension of $V$ is at most 1 (such matrices do not exist hence they have infinite order). 
 A: This is true. Let $M$ be such a $n\times n$ matrix. First notice that all eigenvalues of $M$ are equal to $1$ (if the eigenvector belongs to $W$, it is clear; if it does not, then $\lambda u=u+w$, so $(\lambda-1)u=w$ so $\lambda=1$). Therefore, the only eigenvalue is 1, which has algebraic multiplicity $n$. It can not have geometric multiplicity $n$ since it then would be the identity matrix.
As explained here, any eigenvalue of a $k$-th power of a matrix $A$ is $k$-th power of some eigenvalue of $A$. The algebraic multiplicity of the former is equal to sum of algebraic multiplicites of its $k$-th roots. If the eigenvalue is non-zero, then the geometric multiplicity is also equal to the sum of geometric multiplicities of its $k$-th roots. 
What this means for $M$ is that for any $k\geq 1$, the only eigenvalue of $M^k$ is $1$, it has algebraic multiplicity $n$, and its geometric multiplicity is equal to the geometric multiplicity of $1$ for $M$ (in particular, less than $n$). So no power of $M$ is diagonalizable. In particular, no power of $M$ is the identity. 
